Nakayama’s Lemma

The following is a short statement which has far-reaching applications. Since its main applications are for local rings, we will state the result in this context. Throughout this section, is a fixed local ring.

Suppose is a minimal set of generators of M. If n = 0, M = 0 and we are done. Otherwise, we have so we can write

But since , we have a unit, so in fact:

and M has a smaller set of generators given by (contradiction). ♦

Corollary 1.

Let M be a finitely generated module over and . If span it as a vector space over , then generate it as an A-module.

Proof

Let be the A-linear map which takes . Let C be the cokernel of f so we get an exact sequence of A-modules: . Since the functor is right-exact (by corollary 2 here), we get an exact:

By the given condition is surjective so . Since C is finitely generated (being a quotient of M), we have C = 0. ♦

Hence, for a finitely generated M over , let and pick such that form a basis over . Then form a minimal generating set for M over A. Furthermore, all minimal generating sets of M have the same cardinality: n.

Cotangent Spaces

The cotangent space of A is given by . More generally, for , the r-th order cotangent space of A is .

Exercise B

Find a local ring which is not a field, satisfying .

[ Hint: take the ring of all differentiable functions ℝ → ℝ and consider the maximal ideal of all functions vanishing at 0. ]

Let us assume that is a finitely generated module; hence so is every . The anomaly in exercise B thus does not occur. From Nakayama’s lemma, we see that a minimal generating set of is obtained from a basis of over .

In the following examples, we will let A be the coordinate ring of a variety V over k (algebraically closed field). For , let so that . We can apply our above reasoning to the local ring and . Note that

since the RHS is already a B-module. Thus we only need to compute:

to find the minimum number of generators of .

Example 1

Suppose and . Then and has k-basis given by all monomials of degree r. The number of such monomials is the number of solutions to in non-negative integers. By elementary combinatorics, we have

which is polynomial in r of degree n-1.

In the following examples, when A is a quotient of we will write instead of for its image in A.

Example 2

Let and . Since in A, any monomial of degree r which contains lies in . Thus has basis and

Example 3

Let and . In , the elements do not have any linear relation modulo , hence we have .

Similarly, in the monomials of degree 2 have no linear relation modulo so we have .

For where , we have in A so a basis is given by where and . Thus . In conclusion

Note

We thus see that in all three case, the function can be represented by a polynomial for sufficiently large r. This is called the Hilbert polynomial of the local ring A; we shall have more to say about this later.

Question to Ponder

What can we say about the degree of the Hilbert polynomial?

F. P. + Flat = F. G. + Projective

Finally, here is an interesting result. Recall that saying a module is finite presented is a stronger condition than saying it is finitely generated. On the other hand, by corollary 1 here, projective modules are flat. It turns out these two pairs of properties complement each other.

Proposition 1.

An A-module M is finitely presented and flat if and only if it is finitely generated and projective.

By localization, we may assume is local (exercise: write out the detailed argument). Hence it suffices to show the following. ♦

Proposition 2.

A finitely presented flat module M over a local ring is free.

Proof

Pick a minimal generating set over A. Take the surjective map which maps and let K be the kernel of this map. Hence we get an exact sequence

By the exercise after this, tensoring this exact sequence with any A-module N gives an exact sequence. In particular, we set to obtain:

But since ‘s form a minimal generating set of M over A, form a basis over . Thus the map is an isomorphism and we have . By proposition 1 here, since M is finitely presented, K is finitely generated. Thus by Nakayama’s lemma K = 0 and . ♦

It remains to fill in the gap in the proof.

Exercise C

Suppose we have a short exact sequence of A-modules

where M is A-flat. Then tensoring with any A-module N gives a short exact sequence:

[ Hint: let be a surjective map where F is free; let K be its kernel so we get a short exact sequence . Since M and F are flat, we have a diagram where the rows and columns are exact. Do a diagram-chase. ]